3.99.46 \(\int \frac {e^{-\frac {2 (-5 x+(25+40 x+16 x^2+e^{2 x} x^2+e^x (10 x+8 x^2)) \log ^2(x))}{x}} (e^{\frac {2 (-5 x+(25+40 x+16 x^2+e^{2 x} x^2+e^x (10 x+8 x^2)) \log ^2(x))}{x}} (2 x^2+2 x^3)+(-900-1440 x-576 x^2-36 e^{2 x} x^2+e^x (-360 x-288 x^2)) \log (x)+(450-288 x^2+e^x (-324 x^2-144 x^3)+e^{2 x} (-18 x^2-36 x^3)) \log ^2(x)+e^{\frac {-5 x+(25+40 x+16 x^2+e^{2 x} x^2+e^x (10 x+8 x^2)) \log ^2(x)}{x}} (6 x^2+(-300-780 x-672 x^2-192 x^3+e^x (-120 x-216 x^2-96 x^3)+e^{2 x} (-12 x^2-12 x^3)) \log (x)+(150+150 x-96 x^2-96 x^3+e^x (-108 x^2-156 x^3-48 x^4)+e^{2 x} (-6 x^2-18 x^3-12 x^4)) \log ^2(x)))}{x^2} \, dx\)

Optimal. Leaf size=30 \[ \left (1+3 e^{5-\left (4+e^x+\frac {5}{x}\right )^2 x \log ^2(x)}+x\right )^2 \]

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Rubi [F]  time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x + 8*x^2))*Log[x]^2))/x)*(2*x^2 + 2*x^3) + (-9
00 - 1440*x - 576*x^2 - 36*E^(2*x)*x^2 + E^x*(-360*x - 288*x^2))*Log[x] + (450 - 288*x^2 + E^x*(-324*x^2 - 144
*x^3) + E^(2*x)*(-18*x^2 - 36*x^3))*Log[x]^2 + E^((-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x + 8*x^
2))*Log[x]^2)/x)*(6*x^2 + (-300 - 780*x - 672*x^2 - 192*x^3 + E^x*(-120*x - 216*x^2 - 96*x^3) + E^(2*x)*(-12*x
^2 - 12*x^3))*Log[x] + (150 + 150*x - 96*x^2 - 96*x^3 + E^x*(-108*x^2 - 156*x^3 - 48*x^4) + E^(2*x)*(-6*x^2 -
18*x^3 - 12*x^4))*Log[x]^2))/(E^((2*(-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x + 8*x^2))*Log[x]^2))
/x)*x^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [B]  time = 0.81, size = 61, normalized size = 2.03 \begin {gather*} 9 e^{10-\frac {2 \left (5+\left (4+e^x\right ) x\right )^2 \log ^2(x)}{x}}+6 e^{5-\frac {\left (5+\left (4+e^x\right ) x\right )^2 \log ^2(x)}{x}} (1+x)+x (2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x + 8*x^2))*Log[x]^2))/x)*(2*x^2 + 2*x^3)
+ (-900 - 1440*x - 576*x^2 - 36*E^(2*x)*x^2 + E^x*(-360*x - 288*x^2))*Log[x] + (450 - 288*x^2 + E^x*(-324*x^2
- 144*x^3) + E^(2*x)*(-18*x^2 - 36*x^3))*Log[x]^2 + E^((-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x
+ 8*x^2))*Log[x]^2)/x)*(6*x^2 + (-300 - 780*x - 672*x^2 - 192*x^3 + E^x*(-120*x - 216*x^2 - 96*x^3) + E^(2*x)*
(-12*x^2 - 12*x^3))*Log[x] + (150 + 150*x - 96*x^2 - 96*x^3 + E^x*(-108*x^2 - 156*x^3 - 48*x^4) + E^(2*x)*(-6*
x^2 - 18*x^3 - 12*x^4))*Log[x]^2))/(E^((2*(-5*x + (25 + 40*x + 16*x^2 + E^(2*x)*x^2 + E^x*(10*x + 8*x^2))*Log[
x]^2))/x)*x^2),x]

[Out]

9*E^(10 - (2*(5 + (4 + E^x)*x)^2*Log[x]^2)/x) + 6*E^(5 - ((5 + (4 + E^x)*x)^2*Log[x]^2)/x)*(1 + x) + x*(2 + x)

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fricas [B]  time = 0.78, size = 153, normalized size = 5.10 \begin {gather*} {\left ({\left (x^{2} + 2 \, x\right )} e^{\left (\frac {2 \, {\left ({\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x\right )}}{x}\right )} + 6 \, {\left (x + 1\right )} e^{\left (\frac {{\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x}{x}\right )} + 9\right )} e^{\left (-\frac {2 \, {\left ({\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2+(((-12*x^4-
18*x^3-6*x^2)*exp(x)^2+(-48*x^4-156*x^3-108*x^2)*exp(x)-96*x^3-96*x^2+150*x+150)*log(x)^2+((-12*x^3-12*x^2)*ex
p(x)^2+(-96*x^3-216*x^2-120*x)*exp(x)-192*x^3-672*x^2-780*x-300)*log(x)+6*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)
*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)+((-36*x^3-18*x^2)*exp(x)^2+(-144*x^3-324*x^2)*exp(x)-288*x^2+450)*log
(x)^2+(-36*exp(x)^2*x^2+(-288*x^2-360*x)*exp(x)-576*x^2-1440*x-900)*log(x))/x^2/exp(((exp(x)^2*x^2+(8*x^2+10*x
)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2,x, algorithm="fricas")

[Out]

((x^2 + 2*x)*e^(2*((x^2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x)^2 - 5*x)/x) + 6*(x + 1)*e^(
((x^2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x)^2 - 5*x)/x) + 9)*e^(-2*((x^2*e^(2*x) + 16*x^2
+ 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x)^2 - 5*x)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (9 \, {\left (16 \, x^{2} + {\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left (4 \, x^{3} + 9 \, x^{2}\right )} e^{x} - 25\right )} \log \relax (x)^{2} - {\left (x^{3} + x^{2}\right )} e^{\left (\frac {2 \, {\left ({\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x\right )}}{x}\right )} + 3 \, {\left ({\left (16 \, x^{3} + 16 \, x^{2} + {\left (2 \, x^{4} + 3 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left (4 \, x^{4} + 13 \, x^{3} + 9 \, x^{2}\right )} e^{x} - 25 \, x - 25\right )} \log \relax (x)^{2} - x^{2} + 2 \, {\left (16 \, x^{3} + 56 \, x^{2} + {\left (x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left (4 \, x^{3} + 9 \, x^{2} + 5 \, x\right )} e^{x} + 65 \, x + 25\right )} \log \relax (x)\right )} e^{\left (\frac {{\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x}{x}\right )} + 18 \, {\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)\right )} e^{\left (-\frac {2 \, {\left ({\left (x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{x} + 40 \, x + 25\right )} \log \relax (x)^{2} - 5 \, x\right )}}{x}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2+(((-12*x^4-
18*x^3-6*x^2)*exp(x)^2+(-48*x^4-156*x^3-108*x^2)*exp(x)-96*x^3-96*x^2+150*x+150)*log(x)^2+((-12*x^3-12*x^2)*ex
p(x)^2+(-96*x^3-216*x^2-120*x)*exp(x)-192*x^3-672*x^2-780*x-300)*log(x)+6*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)
*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)+((-36*x^3-18*x^2)*exp(x)^2+(-144*x^3-324*x^2)*exp(x)-288*x^2+450)*log
(x)^2+(-36*exp(x)^2*x^2+(-288*x^2-360*x)*exp(x)-576*x^2-1440*x-900)*log(x))/x^2/exp(((exp(x)^2*x^2+(8*x^2+10*x
)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2,x, algorithm="giac")

[Out]

integrate(-2*(9*(16*x^2 + (2*x^3 + x^2)*e^(2*x) + 2*(4*x^3 + 9*x^2)*e^x - 25)*log(x)^2 - (x^3 + x^2)*e^(2*((x^
2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x)^2 - 5*x)/x) + 3*((16*x^3 + 16*x^2 + (2*x^4 + 3*x^
3 + x^2)*e^(2*x) + 2*(4*x^4 + 13*x^3 + 9*x^2)*e^x - 25*x - 25)*log(x)^2 - x^2 + 2*(16*x^3 + 56*x^2 + (x^3 + x^
2)*e^(2*x) + 2*(4*x^3 + 9*x^2 + 5*x)*e^x + 65*x + 25)*log(x))*e^(((x^2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x
+ 40*x + 25)*log(x)^2 - 5*x)/x) + 18*(x^2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x))*e^(-2*((
x^2*e^(2*x) + 16*x^2 + 2*(4*x^2 + 5*x)*e^x + 40*x + 25)*log(x)^2 - 5*x)/x)/x^2, x)

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maple [B]  time = 0.20, size = 144, normalized size = 4.80

method	result	size
risch	\(x^{2}+2 x +\left (6 x +6\right ) {\mathrm e}^{-\frac {\ln \relax (x )^{2} {\mathrm e}^{2 x} x^{2}+8 x^{2} {\mathrm e}^{x} \ln \relax (x )^{2}+10 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+16 x^{2} \ln \relax (x )^{2}+40 x \ln \relax (x )^{2}+25 \ln \relax (x )^{2}-5 x}{x}}+9 \,{\mathrm e}^{-\frac {2 \left (\ln \relax (x )^{2} {\mathrm e}^{2 x} x^{2}+8 x^{2} {\mathrm e}^{x} \ln \relax (x )^{2}+10 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+16 x^{2} \ln \relax (x )^{2}+40 x \ln \relax (x )^{2}+25 \ln \relax (x )^{2}-5 x \right )}{x}}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+2*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+16*x^2+40*x+25)*ln(x)^2-5*x)/x)^2+(((-12*x^4-18*x^3-
6*x^2)*exp(x)^2+(-48*x^4-156*x^3-108*x^2)*exp(x)-96*x^3-96*x^2+150*x+150)*ln(x)^2+((-12*x^3-12*x^2)*exp(x)^2+(
-96*x^3-216*x^2-120*x)*exp(x)-192*x^3-672*x^2-780*x-300)*ln(x)+6*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+1
6*x^2+40*x+25)*ln(x)^2-5*x)/x)+((-36*x^3-18*x^2)*exp(x)^2+(-144*x^3-324*x^2)*exp(x)-288*x^2+450)*ln(x)^2+(-36*
exp(x)^2*x^2+(-288*x^2-360*x)*exp(x)-576*x^2-1440*x-900)*ln(x))/x^2/exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+16*
x^2+40*x+25)*ln(x)^2-5*x)/x)^2,x,method=_RETURNVERBOSE)

[Out]

x^2+2*x+(6*x+6)*exp(-(ln(x)^2*exp(2*x)*x^2+8*x^2*exp(x)*ln(x)^2+10*x*exp(x)*ln(x)^2+16*x^2*ln(x)^2+40*x*ln(x)^
2+25*ln(x)^2-5*x)/x)+9*exp(-2*(ln(x)^2*exp(2*x)*x^2+8*x^2*exp(x)*ln(x)^2+10*x*exp(x)*ln(x)^2+16*x^2*ln(x)^2+40
*x*ln(x)^2+25*ln(x)^2-5*x)/x)

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maxima [B]  time = 0.55, size = 128, normalized size = 4.27 \begin {gather*} x^{2} + 3 \, {\left (2 \, {\left (x e^{5} + e^{5}\right )} e^{\left (-x e^{\left (2 \, x\right )} \log \relax (x)^{2} + 8 \, x e^{x} \log \relax (x)^{2} + 16 \, x \log \relax (x)^{2} + 10 \, e^{x} \log \relax (x)^{2} + 40 \, \log \relax (x)^{2} + \frac {25 \, \log \relax (x)^{2}}{x}\right )} + 3 \, e^{\left (-2 \, x e^{\left (2 \, x\right )} \log \relax (x)^{2} + 10\right )}\right )} e^{\left (-16 \, x e^{x} \log \relax (x)^{2} - 32 \, x \log \relax (x)^{2} - 20 \, e^{x} \log \relax (x)^{2} - 80 \, \log \relax (x)^{2} - \frac {50 \, \log \relax (x)^{2}}{x}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2+(((-12*x^4-
18*x^3-6*x^2)*exp(x)^2+(-48*x^4-156*x^3-108*x^2)*exp(x)-96*x^3-96*x^2+150*x+150)*log(x)^2+((-12*x^3-12*x^2)*ex
p(x)^2+(-96*x^3-216*x^2-120*x)*exp(x)-192*x^3-672*x^2-780*x-300)*log(x)+6*x^2)*exp(((exp(x)^2*x^2+(8*x^2+10*x)
*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)+((-36*x^3-18*x^2)*exp(x)^2+(-144*x^3-324*x^2)*exp(x)-288*x^2+450)*log
(x)^2+(-36*exp(x)^2*x^2+(-288*x^2-360*x)*exp(x)-576*x^2-1440*x-900)*log(x))/x^2/exp(((exp(x)^2*x^2+(8*x^2+10*x
)*exp(x)+16*x^2+40*x+25)*log(x)^2-5*x)/x)^2,x, algorithm="maxima")

[Out]

x^2 + 3*(2*(x*e^5 + e^5)*e^(-x*e^(2*x)*log(x)^2 + 8*x*e^x*log(x)^2 + 16*x*log(x)^2 + 10*e^x*log(x)^2 + 40*log(
x)^2 + 25*log(x)^2/x) + 3*e^(-2*x*e^(2*x)*log(x)^2 + 10))*e^(-16*x*e^x*log(x)^2 - 32*x*log(x)^2 - 20*e^x*log(x
)^2 - 80*log(x)^2 - 50*log(x)^2/x) + 2*x

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mupad [B]  time = 6.21, size = 121, normalized size = 4.03 \begin {gather*} 2\,x+9\,{\mathrm {e}}^{10-80\,{\ln \relax (x)}^2-\frac {50\,{\ln \relax (x)}^2}{x}-20\,{\mathrm {e}}^x\,{\ln \relax (x)}^2-2\,x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (x)}^2-16\,x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2-32\,x\,{\ln \relax (x)}^2}+{\mathrm {e}}^{5-40\,{\ln \relax (x)}^2-\frac {25\,{\ln \relax (x)}^2}{x}-10\,{\mathrm {e}}^x\,{\ln \relax (x)}^2-x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (x)}^2-8\,x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2-16\,x\,{\ln \relax (x)}^2}\,\left (6\,x+6\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(5*x - log(x)^2*(40*x + x^2*exp(2*x) + exp(x)*(10*x + 8*x^2) + 16*x^2 + 25)))/x)*(exp(-(5*x - log
(x)^2*(40*x + x^2*exp(2*x) + exp(x)*(10*x + 8*x^2) + 16*x^2 + 25))/x)*(log(x)^2*(exp(x)*(108*x^2 + 156*x^3 + 4
8*x^4) - 150*x + exp(2*x)*(6*x^2 + 18*x^3 + 12*x^4) + 96*x^2 + 96*x^3 - 150) + log(x)*(780*x + exp(2*x)*(12*x^
2 + 12*x^3) + 672*x^2 + 192*x^3 + exp(x)*(120*x + 216*x^2 + 96*x^3) + 300) - 6*x^2) - exp(-(2*(5*x - log(x)^2*
(40*x + x^2*exp(2*x) + exp(x)*(10*x + 8*x^2) + 16*x^2 + 25)))/x)*(2*x^2 + 2*x^3) + log(x)^2*(exp(x)*(324*x^2 +
144*x^3) + exp(2*x)*(18*x^2 + 36*x^3) + 288*x^2 - 450) + log(x)*(1440*x + 36*x^2*exp(2*x) + exp(x)*(360*x + 2
88*x^2) + 576*x^2 + 900)))/x^2,x)

[Out]

2*x + 9*exp(10 - 80*log(x)^2 - (50*log(x)^2)/x - 20*exp(x)*log(x)^2 - 2*x*exp(2*x)*log(x)^2 - 16*x*exp(x)*log(
x)^2 - 32*x*log(x)^2) + exp(5 - 40*log(x)^2 - (25*log(x)^2)/x - 10*exp(x)*log(x)^2 - x*exp(2*x)*log(x)^2 - 8*x
*exp(x)*log(x)^2 - 16*x*log(x)^2)*(6*x + 6) + x^2

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sympy [B]  time = 138.09, size = 99, normalized size = 3.30 \begin {gather*} x^{2} + 2 x + \left (6 x + 6\right ) e^{- \frac {- 5 x + \left (x^{2} e^{2 x} + 16 x^{2} + 40 x + \left (8 x^{2} + 10 x\right ) e^{x} + 25\right ) \log {\relax (x )}^{2}}{x}} + 9 e^{- \frac {2 \left (- 5 x + \left (x^{2} e^{2 x} + 16 x^{2} + 40 x + \left (8 x^{2} + 10 x\right ) e^{x} + 25\right ) \log {\relax (x )}^{2}\right )}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+2*x**2)*exp(((exp(x)**2*x**2+(8*x**2+10*x)*exp(x)+16*x**2+40*x+25)*ln(x)**2-5*x)/x)**2+(((-
12*x**4-18*x**3-6*x**2)*exp(x)**2+(-48*x**4-156*x**3-108*x**2)*exp(x)-96*x**3-96*x**2+150*x+150)*ln(x)**2+((-1
2*x**3-12*x**2)*exp(x)**2+(-96*x**3-216*x**2-120*x)*exp(x)-192*x**3-672*x**2-780*x-300)*ln(x)+6*x**2)*exp(((ex
p(x)**2*x**2+(8*x**2+10*x)*exp(x)+16*x**2+40*x+25)*ln(x)**2-5*x)/x)+((-36*x**3-18*x**2)*exp(x)**2+(-144*x**3-3
24*x**2)*exp(x)-288*x**2+450)*ln(x)**2+(-36*exp(x)**2*x**2+(-288*x**2-360*x)*exp(x)-576*x**2-1440*x-900)*ln(x)
)/x**2/exp(((exp(x)**2*x**2+(8*x**2+10*x)*exp(x)+16*x**2+40*x+25)*ln(x)**2-5*x)/x)**2,x)

[Out]

x**2 + 2*x + (6*x + 6)*exp(-(-5*x + (x**2*exp(2*x) + 16*x**2 + 40*x + (8*x**2 + 10*x)*exp(x) + 25)*log(x)**2)/
x) + 9*exp(-2*(-5*x + (x**2*exp(2*x) + 16*x**2 + 40*x + (8*x**2 + 10*x)*exp(x) + 25)*log(x)**2)/x)

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